Der Neupreis des Autos beträgt $37450\text{ €}\mathrm{.}\backslash \; 37450\backslash \; €.$

Die Formel für den exponentiellen Zerfall lautet:

$N\left(t\right)=N_0\cdot (1-p{)}^t\backslash hspace\{35mm\}N\backslash left(t\backslash right)=N\_0\backslash cdot(1-p)^t$

$N(t)=25000N_0=37450p={\displaystyle \frac{\mathrm{12,7}\mathrm{\%}}{100\mathrm{\%}}}=\mathrm{0,127}N(t)\backslash \; =\backslash \; 25000\backslash quad\; N\_0=37450\backslash quad\; p\backslash \; =\backslash \; \backslash dfrac\{12\{,\}7\backslash \%\}\{100\backslash \%\}=0\{,\}127\backslash \; \backslash$setze die Werte in die Formel ein.

$25000=37450\cdot (1-\mathrm{0,127}{)}^t\backslash hspace\{35mm\}25000=37450\backslash cdot(1-0\{,\}127)^\{t\}$

${\displaystyle \frac{25000}{37450}}={\mathrm{0,873}}^t\backslash hspace\{35mm\}\backslash dfrac\{25000\}\{37450\}=0\{,\}873^\{t\}$

${\mathrm{0,873}}^t={\displaystyle \frac{25000}{37450}}\backslash hspace\{35mm\}0\{,\}873^\{t\}=\backslash dfrac\{25000\}\{37450\}$

Berechne $\text{t}\backslash text\{t\}$ mit Hilfe des Logarithmus.

$t={\log }_{\mathrm{0,873}}\left({\displaystyle \frac{25000}{37450}}\right)\backslash hspace\{30mm\}t=\backslash log\_\{0\{,\}873\}\backslash left(\backslash dfrac\{25000\}\{37450\}\backslash right)$

$t={\displaystyle \frac{lg25000-lg37450}{lg\mathrm{0,873}}}\Rightarrow t\approx {\displaystyle \frac{-\mathrm{0,176}}{-\mathrm{0,059}}}\Rightarrow t\approx 3\text{ Jahre}\backslash hspace\{30mm\}t\backslash \; =\backslash \; \backslash dfrac\{lg\backslash \; 25000-lg\backslash \; 37450\}\{lg\backslash \; 0\{,\}873\}\backslash \; \backslash Rightarrow\backslash \; t\backslash \; \backslash approx\backslash \; \backslash dfrac\{-0\{,\}176\}\{-0\{,\}059\}\backslash \; \backslash Rightarrow\backslash \; t\backslash \; \backslash approx\backslash \; 3\backslash \; \backslash text\{Jahre\}$

Nach ungefähr $3\backslash \; 3\backslash$ Jahren verringert sich der Wert des Autos von $37500\text{ € }\backslash \; 37500\backslash \; €\backslash$ auf $25000\text{ €}\backslash \; 25000\backslash \; €$.

Wert des Neuwagens nach $8\backslash \; 8\backslash$ Jahren: $9000\text{ €}\backslash \; 9000\backslash \; €$

Die Formel fur das exponentielle Wachstum lautet:

$N\left(t\right)=N_0\cdot (1-p{)}^t\backslash hspace\{4cm\}N\backslash left(t\backslash right)=N\_0\backslash cdot(1-p)^t$

$N(t)=9000N_0=37450t=8\text{ Jahre}N(t)\backslash \; =\backslash \; 9000\backslash qquad\; N\_0=37450\backslash qquad\; t\backslash \; =\backslash \; 8\backslash \; \backslash text\{Jahre\}\backslash qquad$setze die Werte in die Formel ein.

$9000=37450\cdot (1-p{)}^8\backslash hspace\{4cm\}9000=37450\backslash cdot(1-p)^\{8\}$

${\displaystyle \frac{9000}{37450}}=(1-p{)}^8\backslash hspace\{4cm\}\backslash dfrac\{9000\}\{37450\}=(1-p)^\{8\}$

$(1-p{)}^8={\displaystyle \frac{9000}{37450}}\backslash hspace\{4cm\}(1-p)^\{8\}=\backslash dfrac\{9000\}\{37450\}$

$(1-p)=\sqrt[8]{{\displaystyle \frac{9000}{37450}}}\backslash hspace\{4cm\}(1-p)=\backslash sqrt[\backslash scriptsize\{8\}]\{\backslash dfrac\{9000\}\{37450\}\}$

$1-p=\mathrm{0,837}\Rightarrow -p=\mathrm{0,837}-1\backslash hspace\{3cm\}1-p=0\{,\}837\backslash quad\backslash Rightarrow\backslash quad\; -p=0\{,\}837-1$

$p\approx \mathrm{0,163}\backslash hspace\{50mm\}p\backslash approx0\{,\}163$

Der Wertverlust pro Jahr beträgt ungefähr$\boxed{{\displaystyle \mathrm{16,3}\mathrm{\%}}}\backslash \; \backslash boxed\{16\{,\}3\backslash \%\}$

Die Formel für den exponentiellen Zerfall lautet:

$N\left(t\right)=N_0\cdot (1-p{)}^t\backslash hspace\{4cm\}N\backslash left(t\backslash right)=N\_0\backslash cdot(1-p)^t$

$N_0=37450p_1=25\mathrm{\%}{t}_1=1\text{ Jahre}{p}_2=18\mathrm{\%}{t}_2=1\text{ Jahre}{p}_3=9\mathrm{\%}{t}_3=4\text{ Jahre}N\_0\backslash \; =\backslash \; 37450\backslash quad\; p\_\{1\}\backslash \; =\backslash \; 25\backslash \%\backslash \; \backslash \; t\_\{1\}\backslash \; =\backslash \; 1\backslash \; \backslash text\{Jahre\}\backslash quad\; \backslash \backslash \backslash hspace\{22mm\}p\_\{2\}\backslash \; =\backslash \; 18\backslash \%\backslash \; \backslash \; t\_\{2\}\backslash \; =\backslash \; 1\backslash \; \backslash text\{Jahre\}\backslash quad\; \backslash \backslash \backslash hspace\{22mm\}p\_\{3\}\backslash \; =\backslash \; \backslash \; \backslash \; 9\backslash \%\backslash \; \backslash \; t\_\{3\}\backslash \; =\backslash \; 4\backslash \; \backslash text\{Jahre\}$

Berechne den Wert des Autos nach 6 Jahren.

$N\left(6\right)=37450\cdot (1-\mathrm{0,25}{)}^1\cdot (1-\mathrm{0,18}{)}^1\cdot (1-\mathrm{0,09}{)}^4\backslash \; N\backslash left(6\backslash right)=37450\backslash cdot(1-0\{,\}25)^1\backslash cdot(1-0\{,\}18)^1\; \backslash cdot(1-0\{,\}09)^4$

$N\left(6\right)=37450\cdot {\mathrm{0,75}}^1\cdot {\mathrm{0,82}}^1\cdot {\mathrm{0,91}}^4\backslash \; N\backslash left(6\backslash right)=37450\backslash cdot0\{,\}75^1\backslash cdot0\{,\}82^1\backslash cdot0\{,\}91^4$

$N\left(6\right)=37450\cdot \mathrm{0,421736}N\left(6\right)\approx 15794\text{ €}\backslash \; N\backslash left(6\backslash right)=37450\backslash cdot0\{,\}421736\backslash \backslash \backslash \; N\backslash left(6\backslash right)\backslash approx15794\backslash \; €$

Nach $66$ Jahren beträgt der Wert des Autos noch ungefähr $1579415794$ Euro.